Explanation

(b)
Oxide I Oxide II
Metal, $M$ 50% 40%
Oxygen, $O$ 50% 60%
As first oxide is $MO_{2}$
Let atomic mass of $M = x$
$\therefore$ %$O = \frac{32}{x + 32} \times 100$
Or $\frac{50}{100} = \frac{32}{x + 32}$
Or $0.5 = \frac{32}{x + 32}$
Or $0.5 \times x + 16 = 32$
Or $0.5x = 16$
$x = 32$
$\therefore$At. Mass of metal $M = 32$
Let formula of second oxide is $M_{2}O_{n}$
%$M = \frac{2x}{2x + 16n} \times 100 = \frac{64}{64 + 16n} \times 100$
$\frac{40}{100} = \frac{64}{64 + 16n}$
Or $\frac{100}{40} = \frac{64 + 16n}{64}$
$2.5 = 1 + 0.25\ n$
$n = \frac{1.5}{0.25} = 6$
Therefore, formula of second oxide $= M_{2}O_{6}$
Or $= MO_{3}$